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Complex Analysis

Suppose that $f$ is a holomorphic function in $\mathbb{C}\setminus\{0\}$ and satisfies $$\vert f(z) \vert \leq \vert z \vert ^{2} +\frac{1}{\vert z \vert ^{2}}$$

If f is an odd function, find an expression for f.

I thought about calculating the Laurent Serie of f, but I can't calculate Laurent's series explicitly (there is no way to calculate the coefficients), Dai thought about estimating, using the invariance of the circle to know which one cancels out. I cannot develop this question

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    $\begingroup$ Hint: Show that $z^2f(z)$ is a polynomial of degree $\leq 4$ $\endgroup$ – leoli1 Jan 17 at 14:13
  • $\begingroup$ Which of the terms in the Laurent series must vanish? $\endgroup$ – Vercassivelaunos Jan 17 at 14:18
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The inequality implies that for $|z|<1$ $$ |z^2f(z)|\le |z|^4+1\le 2\ . $$

It follows from Riemann's theorem that $z=0$ is a removable singularity for $g(z):=z^2f(z)$. Thus, $g(z)=\sum_{n=0}^\infty a_nz^n$ on the punctured complex plane.

The polynomial growth $$ |g(z)|\le |z|^4+1 $$ implies that $g$ is a polynomial (or order at most $4$) by Liouville’s theorem

So you can write $$ z^2f(z)=a_0+a_1z+a_2z^2+a_3z^3+a_4z^4. $$

The parity of $f$ then implies that $a_0=a_2=a_4=0$ by the fundamental theorem of algebra: $g(z)+g(-z)=0$ implies that $$ a_0+a_2z^2+a_4z^4=0 $$ for all $z$. (But a nonzero polynomial has only finitely many roots.)

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  • $\begingroup$ I can't understand the fact that "The parity of f then implies that $a_0 = a_2 = a_4 = 0$ by the fundamental theorem of algebra." Could you explain it better? $\endgroup$ – User Jan 17 at 17:52
  • $\begingroup$ @User: f is odd implies (by the definition of "odd") that $f(z)+f(-z)=0$, so $g(z)+g(-z)=0$.// Do you know why $a_0+a_2z^2+a_4z^4=0$? $\endgroup$ – user9464 Jan 17 at 17:56
  • $\begingroup$ yes, because we have to $g(z)+g(-z)=2(a_0+ a_2 z^2 + a_4 z^4 )$. $g(z)+g(-z)=0$, implies $a_0+a_2 z^2+a_4 z^4 =0$. $\endgroup$ – User Jan 17 at 18:04
  • $\begingroup$ @User: good. Note that this is true for all $z$. Do you know the fundamental theorem of algebra? $\endgroup$ – user9464 Jan 17 at 18:09
  • $\begingroup$ the theorem will guarantee that $a_0 + a_2 z^2 + a_4 z^4 = 0$ has a root in $\mathbb{C}$ $\endgroup$ – User Jan 17 at 18:16
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One can immediately see that for all $z\in\Bbb C^\times:=\Bbb C\setminus\{0\},$ we have $$\left|z^2f(z)\right|=|z|^2|f(z)|\le|z|^4+1.$$

As a result, the Laurent series for $g(z):=z^2f(z)$ cannot have any terms of power greater than $4,$ for if there were any, then for sufficiently large $|z|,$ the above inequality would fail to hold. Moreover, we see that $g$ is bounded in the punctured unit disk, so its Laurent series has no terms of negative power, and so is a polynomial of degree no greater than $4.$

Since $f$ is odd, then $g(-z)=(-z)^2f(-z)=z^2f(-z)=-z^2f(z)=-g(z),$ and so $g$ is odd. Hence, all terms of the Laurent series for $g$ have odd power, and so $$g(z)=az^3+bz$$ for some $a,b\in\Bbb C,$ whence $$f(z)=az+\frac{b}{z}.$$


Added: To see why all terms of the Laurent series of $g$ must have odd power, suppose $h$ is an odd function that is holomorphic in $\Bbb C^\times.$ Note that for all $z\in\Bbb C^\times,$ we have $$0=h(z)+h(-z).\tag{$\heartsuit$}$$ Now, write $$h(z)=\sum_{n\in\Bbb Z}b_nz^n,$$ and consider out the expansion for the right-hand side of $(\heartsuit).$

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  • $\begingroup$ I didn't understand why all terms of the Laurent series for g have odd power $\endgroup$ – User Jan 17 at 15:29
  • $\begingroup$ I've added some details to my answer to help you get there. $\endgroup$ – Cameron Buie Jan 17 at 16:04
  • $\begingroup$ What is the need to consider the Laurent Series? Could we consider just one series of power? $\endgroup$ – User Jan 24 at 1:39
  • $\begingroup$ If we knew that $f$ were defined at $0,$ then we could conclude that such a power series were defined. Equivalently, we could conclude that in the Laurent series $$f(z)=\sum_{n\in\Bbb Z}a_nz^n,$$ we have $a_n=0$ for all $n<0,$ and $z=0$ is a removable singularity of $f.$ $\endgroup$ – Cameron Buie Jan 24 at 1:50
  • $\begingroup$ So, we use the laurent series because $f$ is not defined in z=0. It's just it ? $\endgroup$ – User Jan 24 at 1:57

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